Find the integral of the given function w.r.t - x
y=sin2√x2√x
12√x−14sin 2√x+c
I=∫ydx=∫sin2√x2√xdx
Let u=√x,dudx=12√x
⇒du=dx2√x ......(i)
∴I=∫sin2u du(substituting (i))
= ∫1−cos 2u2du(∵cos2a=1−2sin2a)
= ∫12du−12∫cos2udu
Let t=2u⇒dtdu=2⇒du=dt2
∴I=∫12du−12∫cost dt2
= 12u+c1−14∫cos t dt
= 12u+c1−14sin t+c2
= 12u−14sin t+c3
= 12√x−14sin 2u+c3
= 12√x−14sin 2√x+c3